Logarithm

Laws of LogarithmSolved Examples on LogarithmCharacteristic and MantissaProperties of Logarithm Properties Of Monotonocity Of LogarithmLogarithmic Functions GraphLogarithm problems asked in Exams

Introduction to Logarithm

The logarithm of any positive number, whose base is a number, which is greater than zero and not equal to one, is the index or the power to which the base must be raised in order to obtain the given number.

Mathematically: If

If a = 10, then we write log b rather than log₁₀ b. If a = e, we write ln b rather than log_e b. Here, ‘e’ is called as Napier’s base and has numerical value equal to 2.7182. Also, log₁₀ e is known as Napierian constant.

i.e. log₁₀ e = 0.4343

∴ ln b = 2.303 log₁₀ b

Since,

⇒ Important Points

1. log 2 = log₁₀ 2 = 0.3010
2. log 3 = log₁₀ 3 = 0.4771
3. ln 2 = 2.303 log 2 = 0.693
4. ln 10 = 2.303

Laws of Logarithm

Corollary 1: From the definition of the logarithm of the number b to the base a, we have an identity

Which is known as the Fundamental Logarithmic Identity.

Corollary 2: The function defined by

Corollary 3: 

1. If a > 1, then a^x is monotonically increasing. For example,
   \(\begin{array}{l}{{5}^{2.7}}>{{5}^{2.5}},{{3}^{222}}>{{3}^{111}}\end{array} \)
2. If 0 < a < 1, then a^x is monotonically decreasing. For example,
   \(\begin{array}{l}{{\left( \frac{1}{5} \right)}^{2.7}}<{{\left( \frac{1}{5} \right)}^{2.5}},{{\left( 0.7 \right)}^{222}}<{{\left( 0.7 \right)}^{212}}\end{array} \)

Corollary 4:

1. If a > 1, then
   \(\begin{array}{l}{{a}^{-\infty }}=0\end{array} \)
   i.e.
   \(\begin{array}{l}\,\,\,\,\,{{\log }_{a}}0=-\infty\left( if\,a>1 \right)\end{array} \)
2. If 0 < a < 1, then
   \(\begin{array}{l}{{a}^{\infty }}=0\end{array} \)
   i.e.
   \(\begin{array}{l}\,\,\,\,\,{{\log }_{a}}0=+\infty \left( if\,0<a<1 \right)\end{array} \)

Corollary 5:

• \(\begin{array}{l}{{\log }_{a}}b\to \infty ,if\,a>1,b\to \infty\end{array} \)
• \(\begin{array}{l}{{\log }_{a}}b\to -\infty ,if\,0<a<1,b\to \infty\end{array} \)

Remarks

1. ‘log’ is the abbreviation of the word ‘logarithm’.
2. Common logarithm (Brigg’s logarithms). The base is 10.
3. If x < 0, a > 0 and
   \(\begin{array}{l}a\ne 1,\end{array} \)
   then log_a x is an imaginary.
4. \(\begin{array}{l}{{\log }_{a}}1=0\left( a>0,a\ne 1 \right)\end{array} \)
5. \(\begin{array}{l}{{\log }_{a}}a=1\left( a>0,a\ne 1 \right)\end{array} \)
6. \(\begin{array}{l}{{\log }_{\left( 1/a \right)}}a=-1\left( a>0,a\ne 1 \right)\end{array} \)

If

And if,

Solved Examples on Logarithm

Example: 1: Find the value of

Solution:

Here, base tan 45⁰ = 1

∴ log is not defined.

Example: 2: Find the value of

Solution:

Here, base

∴ log is not defined.

Example: 3: Find the value of

Solution:

Since,

Here, base = 1, ∴ log is not defined.

Example: 4: Find the value of

Solution:

Characteristic and Mantissa

The integral part of a logarithm is called the characteristic and the fractional part (decimal part) is called mantissa.

i.e., log N = Integer + Fractional or decimal part (+ve)

⇒ The mantissa of the log of a number is always kept positive.

i.e., if log564 = 2.751279, then 2 is the characteristic and 0.751279 is the mantissa of the given number 564.

And if log0.00895 = -2.0481769 = -2–0.0481769 = (-2-1)+(1-0.0481769) = -3+0.9518231

Hence, -3 is the characteristic and 0.9518231

(not 0.0481769) is mantissa of log 0.00895.

In short, -3+0.951823 is written as

Important Conclusions on Characteristic And Mantissa

• If the characteristics of log N be n, then the number of digits in N is (n+1) (Here, N > 1).
• If the characteristics of log N be –n, then there exists (n-1) number of zeroes after decimal part of N (here, 0 < N < 1).
• If N > 1, the characteristic of log N will be on less than the number of digits in integral part of N.
• If 0 < N < 1, the characteristic of log N is negative and numerically it is one greater than the number of zeroes immediately after the decimal part in N.

For Example:

1. If log 235.68 = 2.3723227. Here, N = 235.68

∴ Number of digits in an integral part of N = 3

⇒ Characteristic of log 235.68 = N -1 = 3 – 1 = 2

2. If

Here, four zeroes immediately after the decimal point in the number 0.0000279 is

Problem: If log 2 = 0.301 and log 3 = 0.477, find the number of digits in 6²⁰.

Solution: Let P = 6²⁰ = (2×3)²⁰

∴ log P = 20 log (2×3) = 20 {log2+log3}

= 20 {0.301+0.477} = 20 × 0.778 = 15.560

Since, the characteristic of log P is 15, therefore the number of digits in P will be 15 + 1, i.e. 16.

Principle Properties of Logarithm

Following are the logarithm rules.

Let m and n be arbitrary positive numbers, be any real numbers, then

1. Log_a (m n) = log_a m + log_a n

In general, log_a (x₁, x₂, x₃,…, x_n) =

Or

where, i = 1, 2, 3, …, n.

2.

3.

4.

5.

6.

7.

8.

9.

Some Additional Logarithm Properties

Logarithm properties are given below.

1. \(\begin{array}{l}{{a}^{{{\log }_{b}}x}}={{x}^{{{\log }_{b}}a}},b\ne 1,a,b,x\end{array} \)
   are positive numbers.
2. \(\begin{array}{l}{{a}^{{{\log }_{a}}x}}=x,a>0,a\ne 1,x>0\end{array} \)
3. \(\begin{array}{l}{{\log }_{{{a}^{k}}}}x=\frac{1}{k}{{\log }_{a}}x,a>0,a\ne 1,x>0\end{array} \)
4. \(\begin{array}{l}{{\log }_{a}}{{x}^{2k}}=2k{{\log }_{a}}\left| x \right|,a>0,a\ne 1,k\in I\end{array} \)
5. \(\begin{array}{l}{{\log }_{{{a}^{^{2k}}}}}x=\frac{1}{2k}{{\log }_{\left| a \right|}}x,x>0,a\ne \pm 1\,\,and\,\,k\in I\tilde{\ }\left\{ 10 \right\}\end{array} \)
6. \(\begin{array}{l}{{\log }_{{{a}^{\alpha }}}}{{x}^{\beta }}=\frac{\beta }{\alpha }{{\log }_{a}}x,x>0,a>0,a\ne 1,\alpha \ne 0\end{array} \)
7. \(\begin{array}{l}{{\log }_{a}}{{x}^{2}}\ne 2{{\log }_{a}}x,a>0,a\ne 1\end{array} \)

Since, domain of

1. \(\begin{array}{l}a_{b}^{log\;a} =\sqrt{a},\,if\,b={a}^{2},a>0,b>0,b\ne 1\end{array} \)
2. \(\begin{array}{l}a_{b}^{log\;a} =a^{2},\,if\,b=\sqrt{a},a>0,b>0,b\ne 1\end{array} \)

Example 1: Solve the equation

Solution:

⇒

⇒

⇒

⇒

∴ 

∴

Example 2: If

Solution:

Since,

and

∴

⇒

∴ x = 10

Example 3: Prove that

Solution:

Since,

=

=

Hence,

Example 4: Prove that

Solution:

=

Now, let

=

=

=

=

= 3 = RHS

Properties Of Monotonocity Of Logarithm

Logarithm with Constant Base

1. \(\begin{array}{l}{{\log }_{a}}x>{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} x>y>0,\,if\,a>1 \\ 0<x<y,\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
2. \(\begin{array}{l}{{\log }_{a}}x<{{\log }_{a}}y\Leftrightarrow \left\{ \begin{matrix} 0<x<y,\,if\,a>1 \\ x>y>0,\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
3. \(\begin{array}{l}{{\log }_{a}}x>p\Leftrightarrow \left\{ \begin{matrix} x>{{a}^{p}},\,if\,a>1 \\ 0<x<{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)
4. \(\begin{array}{l}{{\log }_{a}}x<p\Leftrightarrow \left\{ \begin{matrix} 0<x<{{a}^{p}},\,if\,a>1 \\ x>{{a}^{p}},\,if\,0<a<1 \\ \end{matrix} \right.\end{array} \)

Logarithm with Variable Base

1. log_x a is defined, if
   \(\begin{array}{l}a>0,x>0,x\ne 1\end{array} \)
2. If a > 1, then log_x a is monotonically decreasing in
   \(\begin{array}{l}\left( 0,1 \right)\cup \left( 1,\infty \right)\end{array} \)
3. If 0 < a < 1, then log_x a is monotonically increasing in
   \(\begin{array}{l}\left( 0,1 \right)\cup \left( 1,\infty \right)\end{array} \)

Key Points

1. If a > 1, p > 1, then log_a p > 0
2. If 0 < a < 1, p > 1, then log_a p < 0
3. If a > 1, 0 < p < 1, then log_a p < 0
4. If p > a > 1, then log_a p > 1
5. If a > p > 1, then 0 < log_a p < 1
6. If 0 < a < p < 1, then 0 < log_a p < 1
7. If 0 < p < a < 1, then log_a p > 1

Graphs Of Logarithmic Functions

1. Graph of y = log_a x, if a > 1 and x > 0

2. Graph of y = log_a x, if 0 < a < 1 and x > 0

If the number x and the base ‘a’ are on the same side of the unity, then the logarithm is positive.

• y = log_a x, a > 1, x > 1
• y = log_a x, 0 < a < 1, 0 < x < 1

If the number x and the base a are on the opposite sides of the unity, then the logarithm is negative.

• y = log_a x, a > 1, 0 < x < 1
• y = log_a x, 0 < a < 1, x > 1

3. Graph of

Graphs are symmetrical about Y-axis.

4. Graph of

Graphs are same in both cases i.e., a > 1 and 0 < a < 1.

5. Graph of

6. Graph of

(where [ . ] denotes the greatest integer function)

Since, when

when

when

Important Shortcuts to Answer Logarithm Problems

1. For a non-negative number ‘a’ and
   \(\begin{array}{l}n\ge 2,n\in N,\sqrt[n]{a}={{a}^{1/n}}.\end{array} \)
2. The number of positive integers having base a and characteristic n is
   \(\begin{array}{l}{{a}^{n+1}}-{{a}^{n}}.\end{array} \)
3. Logarithm of zero and negative real number is not defined.
4. \(\begin{array}{l}\left| {{\log }_{b}}a+{{\log }_{a}}b \right|\ge 2,\forall a>0,a\ne 1,b>0,b\ne 1\end{array} \)
5. \(\begin{array}{l}{{\log }_{2}}{{\log }_{2}}\underbrace{\sqrt{\sqrt{\sqrt{\sqrt{…\sqrt{2}}}}}}_{n\,\,times}=-n\end{array} \)
6. \(\begin{array}{l}{{a}^{\sqrt{{{\log }_{a}}b}}}={{b}^{\sqrt{{{\log }_{b}}a}}}\end{array} \)
7. Logarithms to the base 10 are called common logarithms (Brigg’s logarithms).
8. If
   \(\begin{array}{l}x={{\log }_{c}}b+{{\log }_{b}}c,y={{\log }_{a}}c+{{\log }_{c}}a,z={{\log }_{a}}b+{{\log }_{b}}a,\,\,then\,\,{{x}^{2}}+{{y}^{2}}+{{z}^{2}}-4=xyz.\end{array} \)

Practice Problems on Logarithm

Logarithm examples with solutions are given below.

Problem 1: If log 11 = 1.0414, prove that 10¹¹ > 11¹⁰.

Solution:

and log11¹⁰ = 10log11 = 10 × 1.0414 = 10.414

It is clear that, 11 > 10.414

⇒

⇒

Problem 2: If log₂ (x-2)<log₄ (x-2), find the interval in which x lies.

Solution:

Here, x – 2 > 0

⇒ x > 2 ……………… (i)

and

⇒

⇒

⇒

⇒ x-2<1

⇒ x < 3 ……………… (ii)

From equations (i) and (ii), we get

Problem 3: If

where,

(a) 9 (b) 12 (c) 16 (d) 20

Solution:

Option: (c)

=

∴

Hence,

Problem 4: Number of real roots of equation

(a) 0 (b) 1 (c) 2 (d) infinite

Solution: Option (a)

Equation (i) is defined, if x² – 4x + 3 > 0

⇒

⇒ x < 1 or x > 3 ……….(ii)

Equation (i) reduces to

∴ x = 2, 3 ……….(iii)

From equations (ii) and (iii), use get

∴ Number of real roots = 0.

Problem 5:  If

(a) 40 (b) 80 (c) 120 (d) 160

Solution:

Option: (c)

Since,

⇒

∴

Similarly,

and

∴

=

=

=

On comparing, we have  N = 120

Problem 6:  If a = log₁₂ 18, b = log₂₄ 54 then the value of ab +5 (a–b) is

(a) 0 (b) 4 (c) 1 (d) none of these

Solution:

Option: (c)

we have

Putting x = log₂ 3, we have

Problem 7: The value of

(a) 3 (b) 0 (c) 2 (d) 1

Solution:

Option: (a)

Set log₂ 12 = a,

and

Therefore, the given expression

Problem 8:  The solution of the equation

(a) x = 1 (b) x = 4 (c) x = 3 (d) x = e²

Solution:

Option: (c)

log₂ log x is meaningful if x > 1

Since

So the given equation reduces to

⇒

log x > 0 so log x = 1 i.e. x = 3.

Problem 9:  If log_0.5 (x – 1) < log_0.25 (x – 1), then x lies in the interval.

• \(\begin{array}{l}\left( 2,\infty \right)\end{array} \)
• \(\begin{array}{l}\left( 3,\infty \right)\end{array} \)
• \(\begin{array}{l}\left( -\infty ,0 \right)\end{array} \)
  (d) (0, 3)

Solution:

log_0.5 (x – 1) < log_0.25 (x – 1)

Problem 10:  If n = 2002 !, evaluate

Solution:

We have,

=

=

Problem 11:  If x, y, z > 0 and such that

Solution:

Let

⇒

⇒

⇒

⇒

⇒

Problem 12:  Solve: log₃ {5 + 4 log₃ (x – 1)} = 2

Solution:

Clearly, the given equation is meaningful, if x – 1 > 0 and 5 + 4 log₃ (x – 1) > 0

⇒

⇒

⇒

⇒

Now,

log₃ {5 + 4 log₃ (x – 1)} = 2

⇒

⇒

⇒ x – 1 = 3

∴ x=4

Clearly, x = 4 satisfies (i).

Hence, x = 4 is the solution to the given equation.

Problem 13: Solve log₃ (3^x – 8) = 2 – x.

Solution: Clearly, the given equation is meaningful, if

Now,

log₃ (3^x – 8) = 2 – x

⇒

⇒

⇒

⇒

⇒

∴ x = 2

Clearly, 2 > log₃ 8

Hence, x = 2 is the solution of the given equation.

Problem 14: Solve: x^{2 log x} = 10x²

Solution:

Clearly, the given equation is meaningful for x > 0.

Now,

x^{2 log x} = 10x²

⇒

⇒

⇒

⇒

⇒

⇒

⇒

⇒

Problem 15:  Solve: log₂ (9-2^x) = 10^{log (3-x)}

Solution:

We observe that the two sides of the given equation are meaningful, if

9 – 2^x > 0 and 3 – x > 0

⇒

⇒

x < 3 ………. (i)

Now,

log₂ (9-2^x) = 10^{log (3-x)}

⇒

⇒

⇒

⇒

⇒

⇒ y = 8, 1

⇒

⇒ x = 3, 0.

But, x = 3 does no satisfy (i).

Hence, x = 0.

Problem 16:  Solve:

Solution:

The given equation is meaningful for x + 1 > 0 i.e. x > -1

Now,

⇒

⇒

⇒

⇒

∴ y = 2, -1

⇒

⇒

⇒ x = 99, x = -0.9

Problem 17:  Evaluate

Solution:

Let

log x = log(72.3)^1/3

⇒

⇒

⇒

⇒

⇒ x = 4.166

Problem 18:  Evaluate

Solution:

Let

log x =  log (10076)^1/5

⇒

⇒

⇒

⇒

⇒ x = 6.409

Problem 19:  What is logarithm of

Solution:

Here we can write

By using the formula

Problem 20:  Prove that,

Solution:

By solving we get

Let x = 1.333 ….. (i)

10x = 13.3333 ….. (ii)

From equation (i) and (ii), we get

So

Now

Problem 21: If

Solution:

Here by using

⇒

Problem 22:  If

Solution:

By using

Clearly x > 0. Then the given equation can be written as

Problem 23:  Prove that,

Solution:

By multiplying and dividing by

⇒

Example 24: Prove that,

Solution:

Here

Hence, by obtaining the value of y we can prove this.

Let

⇒

⇒

∴

Problem 25: Prove that,

Solution:

As similar to Example 3 we can solve it by using

x = 0.4444 ….. (i)

10x = 4.4444 ….. (ii)

Equation (ii) – Equation (i)

So

Also,

Problem26:  Find the value of

Solution:

We can solve above problem by using

⇒

=

Problem 27:  Find the value of,

Solution: Consider

Therefore by using formula

Let

Problem 28:  If

Solution:

By using

We can obtain the value of k and then by comparing it to

Given,

Also,

So

Comparing by

Problem 29:

Solution:

Problem 30: In the expansion of

Solution:

Problem 31:

Solution:

Problem 32: The coefficient of

Solution:

Problem 33:

Solution:

Problem 34:

Problem 35:

Problem 36: